Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

Show Tags

07 Jul 2013, 06:55

5

This post received KUDOS

1

This post was BOOKMARKED

fozzzy wrote:

If y= (x-1)(x+2), then what is the least possible value of y?

A. -3 B. -9/4 C. -2 D. -3/2 E. 0

Any alternative solutions?

The function is a parabola with a positive "a" coefficient

The roots are x=1 and x=-2, for values between 1 and -2 it will have negative values (E is out)

\(y=x^2+x-2\), you can try to insert values (starting from the least) to see if it could be the answer. Example \(-3=x^2+x-2\) or \(x^2+x+1=0\) =>Impossible \(-\frac{9}{4}= x^2+x-2\) or \(x^2+x+\frac{1}{4}=0\) => \(x=-0.5\) valid => CORRECT

Or approach #2:

Given the roots -2 and 1, the x of the vertex will be the middle point => \(x=-0.5\)

And the least value will be the y coordinate of the vertex (plug \(x=-0.5\) into the equation).
_________________

It is beyond a doubt that all our knowledge that begins with experience.

If y= (x-1)(x+2), then what is the least possible value of y?

A. -3 B. -9/4 C. -2 D. -3/2 E. 0

Any alternative solutions?

\(y= (x-1)(x+2)=x^2+x-2\).

Theory: Quadratic expression \(ax^2+bx+c\) reaches its extreme values when \(x=-\frac{b}{2a}\). When \(a>0\) extreme value is minimum value of \(ax^2+bx+c\) (maximum value is not limited). When \(a<0\) extreme value is maximum value of \(ax^2+bx+c\) (minimum value is not limited).

You can look at this geometrically: \(y=ax^2+bx+c\) when graphed on XY plane gives parabola. When \(a>0\), the parabola opens upward and minimum value of \(ax^2+bx+c\) is y-coordinate of vertex, when \(a<0\), the parabola opens downward and maximum value of \(ax^2+bx+c\) is y-coordinate of vertex.

Examples: Expression \(5x^2-10x+20\) reaches its minimum when \(x=-\frac{b}{2a}=-\frac{-10}{2*5}=1\), so minimum value is \(5x^2-10x+20=5*1^2-10*1+20=15\).

Expression \(-5x^2-10x+20\) reaches its maximum when \(x=-\frac{b}{2a}=-\frac{-10}{2*(-5)}=-1\), so maximum value is \(-5x^2-10x+20=-5*(-1)^2-10*(-1)+20=25\).

Back to the original question: \(y= (x-1)(x+2)=x^2+x-2\) --> y reaches its minimum (as \(a=1>0\)) when \(x=-\frac{b}{2a}=-\frac{1}{2}\).

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

Show Tags

27 Oct 2014, 21:04

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

Show Tags

28 Oct 2014, 12:08

Can we also use b^2 - 4ac to determine the valid minimum value, as thats what I did.. I got 0 as the answer for B and I thought 0 means invalid so went for C. I want some clarification here, first if b^2-4ac=0, it means eq has no real soln?? Or my approach for this question was wrong ! Thanks in advance

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

Show Tags

29 Oct 2014, 01:09

I did it this way The minimum (or max, if the coef is -a) should lie half way between the two roots which are -2 and 1. halfway between that is -3/2. Now put in -3/2 in place of x: (-3/2 -1)(-3/2 +2) = -9/4

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

Show Tags

17 Mar 2016, 20:38

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

Show Tags

04 Jul 2016, 02:58

Bunuel wrote:

fozzzy wrote:

If y= (x-1)(x+2), then what is the least possible value of y?

A. -3 B. -9/4 C. -2 D. -3/2 E. 0

Any alternative solutions?

\(y= (x-1)(x+2)=x^2+x-2\).

Theory: Quadratic expression \(ax^2+bx+c\) reaches its extreme values when \(x=-\frac{b}{2a}\). When \(a>0\) extreme value is minimum value of \(ax^2+bx+c\) (maximum value is not limited). When \(a<0\) extreme value is maximum value of \(ax^2+bx+c\) (minimum value is not limited).

You can look at this geometrically: \(y=ax^2+bx+c\) when graphed on XY plane gives parabola. When \(a>0\), the parabola opens upward and minimum value of \(ax^2+bx+c\) is y-coordinate of vertex, when \(a<0\), the parabola opens downward and maximum value of \(ax^2+bx+c\) is y-coordinate of vertex.

Examples: Expression \(5x^2-10x+20\) reaches its minimum when \(x=-\frac{b}{2a}=-\frac{-10}{2*5}=1\), so minimum value is \(5x^2-10x+20=5*1^2-10*1+20=15\).

Expression \(-5x^2-10x+20\) reaches its maximum when \(x=-\frac{b}{2a}=-\frac{-10}{2*(-5)}=-1\), so maximum value is \(-5x^2-10x+20=-5*(-1)^2-10*(-1)+20=25\).

Back to the original question: \(y= (x-1)(x+2)=x^2+x-2\) --> y reaches its minimum (as \(a=1>0\)) when \(x=-\frac{b}{2a}=-\frac{1}{2}\).

Use derivative: \(y'=2x+1\) --> equate to 0: \(2x+1=0\) --> \(x=-\frac{1}{2}\) --> \(y_{min}=(-\frac{1}{2}-1)(-\frac{1}{2}+2)=-\frac{9}{4}\)

Answer: B.

Hope it's clear.

Can we use derivatives and integration to solve such type of problems? would that a bit advanced as per the Gmat exam,however if it helps i don't see the problem in using them

Can we use derivatives and integration to solve such type of problems? would that a bit advanced as per the Gmat exam,however if it helps i don't see the problem in using them

If you know how to use any of the advanced techniques of course you can use it. Personally I've never seen a GMAT question requiring something like this. For example, I've never seen a geometry question requiring trigonometry, every GMAT geometry question can be solved without it.
_________________

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

Show Tags

04 Jul 2016, 06:03

Bunuel wrote:

rhine29388 wrote:

Can we use derivatives and integration to solve such type of problems? would that a bit advanced as per the Gmat exam,however if it helps i don't see the problem in using them

If you know how to use any of the advanced techniques of course you can use it. Personally I've never seen a GMAT question requiring something like this. For example, I've never seen a geometry question requiring trigonometry, every GMAT geometry question can be solved without it.

Always a pleasure learning from you kudos to your post

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

Show Tags

05 Jul 2016, 10:52

1

This post received KUDOS

i did NOT do this problem like the other posters did on here...way too complicated. i just plugged in #s.

start w/ x=1. if x=1, (1-1)=0, so A/C= 0. x=2: (2-1)(4)>0, so 1 is greatest positive # we can do. ok. x=-1: (-2)(1) = -2. Now this is lowest #. Elim 0. x=-2: (-3)(0) = 0. Not quite. x=-3: (-4)(-1) = positive. Not going in right direction. **Answer must be -2<x<0. x= -1/2: (-1/2-2/2)(-1/2+4/2)= (-3/2)(3/2)= -9/4. Good! x= -3/2: (-3/2-2/2)(-3/2+4/2)= (-5/2)(1/2) = -5/4. * -9/4 = -2.25; -5/4= -1.25. -9/4 = correct. A/C B = correct.

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

Show Tags

20 Jul 2017, 06:59

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________